Question

Can you take the natural log of a negative number?

Answer 1

If you are only working in real numbers, no, you cannot take the natural log of a negative number. If you are working in complex (imaginary) numbers, however, it is possible. Explanation below.

Explanation:

Notice that `e^(i pi) = -1` so in some sense you might like to define:

`ln(-1) = pi i`.

Combining this with `ln(ab) = ln(a) + ln(b)` you might then define:

`ln(-x) = ln(x) + pi i` when `x > 0`

However, notice that with this definition:

`ln(-1) + ln(-1) = ln(1) + pi i + ln(1) + pi i = 2 pi i`

Whereas:

`ln(-1 * -1) = ln(1) = 0`

So the `ln(ab) = ln(a) + ln(b)` rule does not quite work for Complex logarithms.

So what's going on?

The problem is that while `e^x:RR->(0,oo)` is a one-one function as a Real valued function, the function `e^z:CC -> CC "\" {0}` is many to one.

So `e^z` does not have a well defined inverse `ln(z):CC "\" {0}->CC` unless we restrict the domain of `e^z` or the range of `ln(z)`.

We can define a principal Complex logarithm consistent with the most common definition of `Arg(z) in (-pi, pi]` as follows:

`ln(r(cos theta + i sin theta)) = ln(r) + i theta`

where `r > 0` and `theta in (-pi, pi]`

This principal Complex logarithm satisfies:

`e^(ln(z)) = z` for any `z in CC "\" {0}`

`ln(e^z) = z + 2pi k i` for some integer `k` for any `z in CC`

Answer 2

If you are restricted to Real numbers, then "no".
If you are working with Complex numbers see George C.'s response.

Explanation:

[I've included this answer only because of uncertainty of the mathematics level at which the question was asked.]

If you could take the (natural) log of a negative number (say `m<0`)
then
`color(white)("XXX")lnm=k`
would mean
`color(white)("XXX")e^k=m < 0`

Since `e~~2.7>0`
`color(white)("XXX")e^k > 0` for all Real values of `k`